Elementary Statistics: Picturing the World (6th Edition)

Published by Pearson
ISBN 10: 0321911210
ISBN 13: 978-0-32191-121-6

Chapter 3 - Probability - Section 3.4 Additional Topics in Probability and Counting - Exercises - Page 174: 13

Answer

$\frac{_{6}P_{2}}{_{11}P_{3}}=\frac{1}{33}\approx0.0303$

Work Step by Step

$_{n}P_{r}=\frac{n!}{(n-r)!}$ $_{6}P_{2}=\frac{6!}{(6-2)!}=\frac{6\times5\times4!}{4!}=6\times5=30$ $_{11}P_{3}=\frac{11!}{(11-3)!}=\frac{11\times10\times9\times8!}{8!}=11\times10\times9=990$ $\frac{_{6}P_{2}}{_{11}P_{3}}=\frac{30}{990}=\frac{1}{33}\approx0.0303$
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